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Math Help - Integration with Euler's constant and cosine

  1. #1
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    Integration with Euler's constant and cosine

    Integrate

    e^(2x)cosx dx

    I tried substitution, where u=2x, but then I am still left with an x value from cosx. I know how to integrate them separately, but now when they are multipled together... how should I go about integrating?
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    How's your "by parts"? Do it twice.
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  3. #3
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    Quote Originally Posted by bhuang View Post
    Integrate

    e^(2x)cosx dx

    I tried substitution, where u=2x, but then I am still left with an x value from cosx. I know how to integrate them separately, but now when they are multipled together... how should I go about integrating?
    You need to use integration by parts.

    \int{u\,dv} = u\,v - \int{v\,du}.


    So for \int{e^{2x}\cos{x}\,dx}

    Let u = e^{2x} so that du = 2e^{2x}.

    Let dv = \cos{x} so that v = \sin{x}.


    Therefore \int{e^{2x}\cos{x}\,dx} = e^{2x}\sin{x} - \int{2e^{2x}\sin{x}\,dx}

     = e^{2x}\sin{x} - 2\int{e^{2x}\sin{x}\,dx}.


    Now you need integration by parts again.

    Let u = e^{2x} so that du = 2e^{2x}.

    Let dv = \sin{x} so that v = -\cos{x}.


    So e^{2x}\sin{x} - 2\int{e^{2x}\sin{x}\,dx} = e^{2x}\sin{x} - 2\left[-e^{2x}\cos{x} - \int{-2e^{2x}\cos{x}\,dx}\right]

     = e^{2x}\sin{x} + 2e^{2x}\cos{x} - 4\int{e^{2x}\cos{x}\,dx}.



    So we have

    \int{e^{2x}\cos{x}\,dx} = e^{2x}\sin{x} + 2e^{2x}\cos{x} - 4\int{e^{2x}\cos{x}\,dx}

    5\int{e^{2x}\cos{x}\,dx} = e^{2x}\sin{x} + 2e^{2x}\cos{x}

    \int{e^{2x}\cos{x}\,dx} = \frac{1}{5}e^{2x}\sin{x} + \frac{2}{5}e^{2x}\cos{x} + C.
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